Question

When 3.06 g of solid $N{H}_4$HS is introduced into a two-litre evacuated flask at 27 ${}^{o}C$, 30$\%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. Choose the most appropriate one regarding $K_c$, $K_p$ and degree of dissociation $\alpha$:

• A. , atm;
• B. , ; is not defined
• C. , ;
• D. and are not defined. But

Answer 1

The correct option is C

, ;

Molar mass (mass of one mole) of $N{H}_{4}SH$ = 51.111 g/mol
So we have a solid dissociating into two gases. Only a part of the solid gets converted into the gases. At equilibrium, the remaining solid is in dynamic, chemical equilibrium with the gases. All these, the solid and the two gases are in equilibrium inside a closed, 2-litre vessel. All this happens at a fixed temperature of 300K.
Clearly, the number of moles of solid at t=0 (at the instant of being introduced) is $\frac{3.06}{51.111}=0.06mol$
Let us tabulate the moles for the reaction:

$\begin{array}{cccccc}& N{H}_{4}HS\left(s\right)& \leftrightharpoons & N{H}_3\left(g\right)& +& H_{2}S\left(g\right)\\ t=0& 3.06g(=0.06mol)& & 0& & 0\\ t_{eq}& 0.7\times 0.06mol& & 0.3\times 0.06mol& & 0.3\times 0.06mol\\ & & & =0.018mol& & =0.018mol\end{array}$
It is given that $30\%$ of the solid is consumed to establish the equilibrium. Let us recall the law of chemical equilibrium. Now I am not going to state it here.
$K_c=\frac{\left[N{H}_3\right]\left[H_{2}S\right]}{\left[N{H}_{4}SH\right]}$
Look at the above equation and think about the states of the compounds. What are the implications?
Volume of the container = 2 L
$K_c$ =$\frac{\frac{0.018}{1}\frac{0.018}{2}}{\frac{0.042}{2}}$ = $3.86\times {10}^{-2}mol{L}^{-1}$
$K_p=K_c\times (R.T{)}^{△n}$
Here $△n$ = 2 - 0 = 2
$K_p$ = $(3.86\times {10}^{-2})(0.0821\times 300{)}^2=23.4162atm$
All this looks great. Except!! The active mass of a solid is 1!! So in that Kc term, $\left[N{H}_{4}SH\right]$ = 1
This changes everything. $K_c$ = $\frac{0.018}{2}\times \frac{0.018}{2}=8.1\times {10}^{-5}$ $(mol/L{)}^2$

Even the unit of Kc changes! Similarly $K_p$ = $K_p$ = $K_c\times (R.T{)}^{△n}$ = $(8.1\times {10}^{-5})(0.0821\times 300{)}^{2}at{m}^2$
$K_p=4.9\times {10}^{-2}at{m}^2$
If we had taken the partial pressure route to arriving at $K_p$, then the previous error in $K_p$ could have been avoided. Why?
Now that the system is in equilibrium, what would happen if we introduce more $N{H}_{4}SH$ (s) into the container? What are its implications on the equilibrium constants?
All this while, the temperature is still 300K. So the values of $K_c$ or $K_p$ would not change at all. Looking at the mathematical expressions for the equilibrium constants, we can say that even the partial pressures of gases will NOT change if more $N{H}_{4}SH$ is introduced.
What is $\alpha$ ? If we define the degree of dissociation as the fraction of the substance present in the dissociated form, then here $\alpha$ = 0.3. But usually, we talk of $\alpha$ in homogenous equilibrium where gases are involved.

Hence the most appropriate solution is c)